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Anettt [7]
3 years ago
10

A car accelerates from 10.0 m/s to 30.0 m/s at a rate of 3.00 m/s2. how far does the car travel while accelerating?

Physics
1 answer:
Snezhnost [94]3 years ago
3 0
Answer is in attachment.

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LuckyWell [14K]
B. <span>much precipitation </span>
3 0
3 years ago
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CAN SOMEONE HELP ME PLEASE!? After chasing its prey, a cougar leaves skid marks that are 236 m in length. Assuming the cougar sk
malfutka [58]

Answer:

u=36.8m/s

Explanation:

because of the acceleration is a constant acceleration we can use one of the "SUVAT" equations

u^2=v^2-2ā*s. where:

u^2 stands for intial velocity

v^2 stands for final velocity

since the cougar skidded to a complete stop the final velocity is zero.

u^2=v^2-2ā*s

u^2=(0)^2 -2(-2.87 m/s^2)*236 m

u^2=0+5.74m/s^2* 236m

u^2=1354.64m^2/s^2

u=√1354.64m^2/s^2

u=36.8m/s (approximate value)

when ever the acceleration is constant you can use one of the following equation to find the required value.

1. v = u + at. (no s)

2. s= 1/2(u+v)t. (no ā)

3. s=ut + 1/2at^2. ( no v)

4. v^2=u^2 + 2āS. (no t). 5. s= vt - 1/2at^2. (no u)

5 0
3 years ago
he fan blades on a jet engine make one thousand revolutions in a time of 54.9 ms. What is the angular frequency of the blades?
Gnesinka [82]

So, the angular frequency of the blades approximately <u>36.43π rad/s</u>.

<h3>Introduction</h3>

Hi ! Here I will discuss about the angular frequency or what is also often called the angular velocity because it has the same unit dimensions. <u>Angular frequency occurs, when an object vibrates (either moving harmoniously / oscillating or moving in a circle)</u>. Angular frequency can be roughly interpreted as the magnitude of the change in angle (in units of rad) per unit time. So, based on this understanding, the angular frequency can be calculated using the equation :

\boxed{\sf{\bold{\omega = \frac{\theta}{t}}}}

With the following condition :

  • \sf{\omega} = angular frequency (rad/s)
  • \sf{\theta} = change of angle value (rad)
  • t = interval of the time (s)

<h3>Problem Solving</h3>

We know that :

  • \sf{\theta} = change of angle value = 1,000 revolution = 1,000 × 2π rad = 2,000π rad/s >> Remember 1 rev = 2π rad/s.
  • t = interval of the time = 54.9 s.

What was asked :

  • \sf{\omega} = angular frequency = ... rad/s

Step by step :

\sf{\omega = \frac{\theta}{t}}

\sf{\omega = \frac{2,000 \pi}{54.9}}

\boxed{\sf{\omega \approx 36.43 \pi \: rad/s}}

<h3>Conclusion :</h3>

So, the angular frequency of the blades approximately 36.43π rad/s.

8 0
2 years ago
An initially uncharged 3.47-μF capacitor and a 6.43-kΩ resistor are connected in series to a 1.50-V battery that has negligible
harkovskaia [24]

Answer: a) io=233.28 A ( initial current); b) τ=R*C= 22.31 ms; c) 81.7 ms

Explanation:  In order to explain this problem we have to use, the formule for the variation of the current in a RC circuit:

I(t)=io*Exp(-t/τ)

and also we consider that io=V/R=(1.5/6.43*10^3)

=233.28 A

then the time constant for the RC circuit is τ=R*C=6.43*10^3*3.47*10^-6

=22.31 ms

Finally the time to reduce the current to 2.57% of its initial value is obtained from:

I(t)=io*Exp(-t/τ)  for I(t)/io=0.0257=Exp(-t/τ) then

ln(0.0257)*τ =-t

t=-ln(0.0257)*τ=81.68 ms

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Which factor is most important in determining climate?
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